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Sum of Terms

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by EmilKhalikov » Sun Dec 21, 2008 5:48 pm
a1 + a2 + ... + aN = 350
...7 + ...7 + ... + ...7 = 350 (n terms)
(...0 + 7) + (...0 + 7) + ... + (...0 + 7) = 350
...0 + ... + ...0 + 7*N = 350

So 7*N = ...0
N could be 40
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by hypik21 » Sun Dec 21, 2008 11:22 pm
divide 350 by 77
comes out to 4 and a remainder of 42= 6 7's , but we're looking for n being near 35-45

so if for every 77, there are 11 7's

1 77 means that weve taken out 3 77's, that ='s 33 7's...plus the 6 from earlier, 39 7's and 1 77 ='s 40 terms
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by amitabhprasad » Sun Dec 21, 2008 11:34 pm
Scan through answer choice to see the range
350/7 = 50
350-77 = 273 = 39*7

==> n = 39+1 = 40
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by mbhusal76 » Tue Dec 23, 2008 4:37 am
Well,
I just used algebraic method.
77x+7y = 350

7x.11 + 7y = 350
7 (11x + 7y) = 350
11x +y = 50

for x =1, y = 50-11 =39
is 39+1 one of the answer.. yes
So answer is C- 40
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by chiller » Wed Jan 14, 2009 5:58 am
hey mbhusal76,

I got as far as you did to get 11x+y=50. However, at that point I don't know how you chose x=1, and y=39. That makes sense; however, if i set x=4, then i only get y=6 - which isn't a choice.

Any ideas?
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